Question

A reminder of the henderson- equation Do the calculation to make 200ml, of 0.1M sodium aculate buffer with a pathe of 5.1 with 5 xm acis and 1 CM NaCH. Calculate the total amount of acetric acid needed. Calculate the radio of the two form of that will when the pH is 5.1 Use this ratio to calculate the % of calculate that will be in the A form. Assume that each NaOH will convert one. Use this plus the % A to calculate the amount of NaOH needed to convert the amount of HAc to How much acetic acid is needed?

Answer 1

200 ml 0.1 M sodium acetate buffer pH 5.1 5 M CH COOH1MNaOH moles of sodium acetate 0.1 Mx 0.2 L =0.02 [Salt] pH -pKa +log [Acid] Ka = 1.8x10-5 1 4.74+log salt] pKa-- log Ka-4.74 acid] [salt] acid] 0.36 = log [salt] acid] 0.36- [salt]- acid] =229 moles ratio since we know the moles of sodium acetate (salt) .2 2.29 acid [acid] = 8.733 x 10-, mole of acetic acid NaOH CH COOH CH COONa (0.02) (0.02) (0.02) Thus. Total moles of CH3COOH 0.028.733 x 100.028733 moles of acetic acid SM CH3COOH 0.005 moles per ml 0.028733 0.0055.74 ml 1 M NaOH 0.001 moles per ml 0.02/0.001 - 20 ml 5.74 ml 5M CH3COOH, 20 ml1 M NaOH and then dilute it to 200 ml